Let f(x) be a function defined in `1lexltoo` by.
`f(x)={:[(2-x,"for "1lexle2),(3x-x^(2),"for "xgt2):}`
Consider the function statements:
I. The function is continuous every point in the interval `[1,oo)`
II. The function is differentiable at `x=15`.
Which of the above statements is/are correct?

Options

A. Only I
B. Only II
C. Both I and II
D. Neither I nor II

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} 2 - x & \text{for } 1 \leq x \leq 2 \\ 3x - x^2 & \text{for } x > 2 \end{cases} \]

We will evaluate the two statements provided:

### Statement I: The function is continuous at every point in the interval \([1, \infty)\).

1. **Check continuity at \( x = 2 \)**:
- We need to find the left-hand limit and the right-hand limit at \( x = 2 \).
- **Left-hand limit** as \( x \) approaches 2:
\[ \lim_{x \to 2^-} f(x) = 2 - 2 = 0 \]
- **Right-hand limit** as \( x \) approaches 2:
\[ \lim_{x \to 2^+} f(x) = 3(2) - (2)^2 = 6 - 4 = 2 \]
- Since the left-hand limit (0) does not equal the right-hand limit (2), the function is not continuous at \( x = 2 \).

2. **Conclusion for Statement I**:
- The function is not continuous at \( x = 2 \), hence it is not continuous on the entire interval \([1, \infty)\).
- Therefore, Statement I is **false**.

### Statement II: The function is differentiable at \( x = 15 \).

1. **Check differentiability at \( x = 15 \)**:
- Since \( 15 > 2 \), we will use the expression \( f(x) = 3x - x^2 \).
- This is a polynomial function, which is differentiable everywhere in its domain.

2. **Conclusion for Statement II**:
- Since \( f(x) = 3x - x^2 \) is differentiable for all \( x > 2 \), it is certainly differentiable at \( x = 15 \).
- Therefore, Statement II is **true**.

### Final Conclusion:
- Statement I is false, and Statement II is true. Thus, the correct answer is that only Statement II is correct.

### Summary of Results:
- **Statement I**: False
- **Statement II**: True